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Posted By: Frank Cinker Little League Baseball Field - 09/11/02 08:27 PM
Please check my math:

(22) 1500 watt metal halide light standard fixtures @208 volts, 8.2 amps each. They will be balanced on a 120/208 volt, three phase, 4 wire service.

37,523 watt load divided by 208 x 1.73=104 amps.
The load would be considered continuous thus multiply by 1.25 = 130 amps total load.

Am I correct?
Posted By: Tom Re: Little League Baseball Field - 09/11/02 08:43 PM
I dunno.

Try this, 21 fixtures installed so they balance out on all 3 phases equals 114.8 amps, add one more fixture on 2 phases brings the total to 123 amps multiply by 1.25 = 153.75 amps

you'll either need a 150 amp panel (your calculation) or a 200 amp panel (my calculation). The price difference is probably non-existant. The only cost that might matter would be the feeder size.
Posted By: tsolanto Re: Little League Baseball Field - 09/11/02 09:29 PM
Frank, I did the calculation and came up with the same amperage as you did. I beleive you are correct
Posted By: electric-ed Re: Little League Baseball Field - 09/11/02 09:36 PM
Each fixture - 208v x 8.2a = 1705.6 va
22 fixtures + 2 (to balance load) = 24 fixtures
1705.6 va x 24 = 40 934.4 va

40 934.4 va divide by 208 = 196.8 amps
divide by 1.73 = 113.7 amps
113.7 x 1.25 (continuous load) = 142.5 amps

Ed
Posted By: Redsy Re: Little League Baseball Field - 09/12/02 03:20 PM
Frank,

Don't forget to add (10%?) for ballasts.
Posted By: Frank Cinker Re: Little League Baseball Field - 09/12/02 07:05 PM
Is that for power factor?
Posted By: Tom Re: Little League Baseball Field - 09/12/02 10:07 PM
Am I missing something? Why is everyone trying to change this calculation to volt-amps when Frank has stated that each fixture draws 8.2 amps?

Doing the Volt-Amp thing might be advantageuos if this load could be equally shared by all three phases, but since it can't, I feel the resulting answers are too low. The math is correct, so I'll assume that everyone slipped off their boots to use their toes [Linked Image] and they all did their gazintas correctly.

There will be 15 fixtures on A phase x 8.2 = 123 amps, 15 fixtures on B phase x 8.2 amps = 123 amps

C phase only has 14 fixtures x 8.2 amp=114.8 amp.

125 amps x 125% = 153.75 amps, too much for 150 amp feeder or 150 amp panel.
Posted By: Chris Rudolph Re: Little League Baseball Field - 09/13/02 12:01 AM
A 1500 watt fixture at 208 volts will draw 7.2amps continuous.Why are we talking 8.2 amps and adding 125% for a continuous load?
What am I missing?
Chris
Posted By: HotLine1 Re: Little League Baseball Field - 09/13/02 12:39 AM
Chris:
A 1500 watt metal halide sport light draws 8.2 amps at 208 voltsat full illumination, and 4.70 amps starting. Source of info is Lumark/Cooper Lighting.

That is a high power factor, CWA.

HotLine1
John

PS to Frank: Put in a 200 amp feeder and panel please!
Posted By: Chris Rudolph Re: Little League Baseball Field - 09/13/02 12:58 AM
John,
Thanks for the info.Does Lumark/Cooper have a WEB site with tech info for their products?
Chris
Posted By: electric-ed Re: Little League Baseball Field - 09/13/02 02:38 AM
Tom,
Frank said he has 22 fixtures, not 44.

Ed
Posted By: resqcapt19 Re: Little League Baseball Field - 09/13/02 02:51 AM
Ed,
What Tom did is put fixture 1 on phases A&B, fixture 2 on B&C, fixture 3 on A&C, and so on. This will give you 15 connections to A and B and 14 connections to C.
Don
Posted By: electric-ed Re: Little League Baseball Field - 09/13/02 03:04 AM
The current drawn by a fixture on phases A and B is not "in phase with" the current drawn by a fixture on phases B and C. They won't both be at "maximum" at the same time. They add, but it would be vectorially.

That's why I would recommend the volt-amp calculation.

ED

[This message has been edited by electric-ed (edited 09-12-2002).]
Posted By: Tom Re: Little League Baseball Field - 09/13/02 11:37 AM
Ed,

I don't quite see the point about about fixtures on A & B not being in phase with fixtures on B & C. The load on B phase will be twice the load on either A or C phase if you have one fixture connected A&B and one fixture connected B&C.

Working with volt-amps may be meaningful for calculating a transformer load, but in this case, I think the higher number is the correct one to use.

The use of amps in the calculation may be a code requirement if I'm reading things right. In the following order 220.10, 220.3(B)(2) .

Of course, I could be just plain stubborn here, after all, that's in the job description of "electrician." [Linked Image]

Tom

[This message has been edited by Tom (edited 09-13-2002).]
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